Question

过椭圆交点的直线与椭圆联立方程消去x还是y简便

Answer 1

右交点坐标F(4,0)，故设直线方程位y=kx+（-4k)=kx-4k设交点A(x1,y1)B(x2,y2)又因为P（0,-4k）PA向量=(x1,y1+4k)=k1AF向量=k1(4-x1,-y1)PB向量=（x2,y2+4k）=k2BF向量=k2(4-x2,-y2)所以k1+k2=x1/(4-x1)+x2/(4-x2)=(4(x1+x2)-2x1x2)/（4-x1)(4-x2)联立椭圆方程和直线方程得(9+25k^2)x^2-200k^2x+25(16k^2-9)=0由韦达定理x1+x2=200k^2/(9+25k^2)x1x2=25(16k^2-9)/(9+25k^2)化解后K1+K2=-50/9满意请采纳

Answer 2

都方便